Question

The value of $\underset{x\to \infty }{lim}{\left(\frac{p^{1/x}+q^{1/x}+r^{1/x}+s^{1/x}}{4}\right)}^{3x},$ where $p,q,r,s>0$ is equal to

• A. $pqrs$
• B. $(pqrs{)}^3$
• C. $(pqrs{)}^{3/2}$
• D. $(pqrs{)}^{3/4}$

Answer 1

The correct option is D $(pqrs{)}^{3/4}$

$L=\underset{x\to \infty }{lim}{\left(\frac{p^{1/x}+q^{1/x}+r^{1/x}+s^{1/x}}{4}\right)}^{3x},$ where $p,q,r,s>0$
Put $\frac{1}{x}=z$
As $x\to \infty ,z\to 0$
$L=\underset{z\to 0}{lim}{\left(\frac{p^z+q^z+r^z+s^z}{4}\right)}^{3/z}$
which is in the form of $1^{\infty }.$

$L=e^{\underset{z\to 0}{lim}(\frac{p^z+q^z+r^z+s^z}{4}-1)\times \frac{3}{z}}$
$=e^{\underset{z\to 0}{lim}(\frac{p^z-1}{z}+\frac{q^z-1}{z}+\frac{r^z-1}{z}+\frac{s^z-1}{z})\times \frac{3}{4}}$
$=e^{3/4({\log }_{e}p+{\log }_{e}q+{\log }_{e}r+{\log }_{e}s)}$
$=e^{3/4{\log }_e\left(pqrs\right)}$
$=(pqrs{)}^{3/4}$